Steel Phantom is a roller coaster in Pennsylvania that, like the Desperado in Nevada, has a vertical drop of 68.6m Suppose a roller-coaster car with a mass of 1000 kg travels from the top of that drop without friction. The car then decelerates along a horizontal stretch of track. If the car travels a distance of 30m along this horizontal track before the car comes to rest, what is the magnitude of its change in mementum? Assume the car's speed at the peak of the drop is zero. What is the magnitude of the average force acting on the car during its deceleration?
Change in momentum = 1000 kg * 68.6 m/s
Average force = Change in momentum / Distance = 1000 kg * 68.6 m/s / 30 m = 2286.67 N
To find the magnitude of the change in momentum, we need to calculate the initial momentum and the final momentum, and then calculate the difference between them.
First, let's calculate the initial momentum.
The initial velocity of the car at the top of the drop is zero, as stated in the question. Therefore, the initial momentum is also zero.
Next, let's calculate the final momentum.
The final velocity of the car when it comes to rest is also zero, as stated in the question. Thus, the final momentum is also zero.
The magnitude of the change in momentum is given by the equation:
Change in momentum = final momentum - initial momentum
Substituting the values, we get:
Change in momentum = 0 - 0 = 0
Therefore, the magnitude of the change in momentum is 0.
Now, let's calculate the magnitude of the average force acting on the car during its deceleration.
We know that force is related to change in momentum by the equation:
Force = change in momentum / time
Since the car comes to rest over a distance of 30m, we can find the time it takes using the equation of motion:
vf^2 = vi^2 + 2ad
Where:
vf = final velocity (0 m/s)
vi = initial velocity (unknown)
a = acceleration (negative, since it's decelerating)
d = distance (30m)
Rearranging the equation:
0 = vi^2 + 2ad
-2ad = vi^2
-2 * a * 30 = vi^2
Given that the acceleration is determined by the change in velocity over the distance, we can use the equation:
a = vf^2 / (2 * d)
Substituting the values:
a = 0 / (2 * 30) = 0
So, the acceleration is zero. Therefore, the magnitude of the average force acting on the car during its deceleration is also zero.
In conclusion, the magnitude of the change in momentum is 0, and the magnitude of the average force acting on the car during its deceleration is also 0.
To answer these questions, we need to consider the conservation of momentum and the work-energy principle.
First, let's find the speed of the roller-coaster car at the bottom of the drop using the conservation of energy. The potential energy at the top of the drop is fully converted into kinetic energy at the bottom, since there is no friction.
Given:
Vertical drop height (h) = 68.6 m
Mass of the car (m) = 1000 kg
Acceleration due to gravity (g) = 9.8 m/s^2
Using the conservation of energy, the potential energy (PE) is converted to kinetic energy (KE), expressed as:
PE = KE
mgh = (1/2)mv^2
Substituting the values, we can solve for v, the speed of the car at the bottom of the drop:
1000 kg * 9.8 m/s^2 * 68.6 m = (1/2) * 1000 kg * v^2
v^2 = 2 * 9.8 m/s^2 * 68.6 m
v^2 = 13418.96 m^2/s^2
v ≈ 115.83 m/s (rounded to two decimal places)
Next, let's calculate the change in momentum (Δp) of the car. The change in momentum is given by:
Δp = mv - mu
Since the car's speed at the top of the drop is zero (u = 0), we have:
Δp = mv - mu = m * (115.83 m/s) - m * 0
Δp = 1000 kg * 115.83 m/s
Δp = 115,830 kg m/s
So, the magnitude of the change in momentum is 115,830 kg m/s.
Now, let's calculate the average force (F) acting on the car during its deceleration along the horizontal track. The work-energy principle states that the net work done on an object is equal to the change in its kinetic energy.
The work done on the car is given by:
Work = Force * Distance
We know that the work done on the car is equal to the change in kinetic energy:
Work = ΔKE
Since the car comes to rest (final kinetic energy is zero), the initial kinetic energy is equal to the work done:
1/2 * m * (v^2) = Force * Distance
Substituting the values:
Force * 30 m = 1/2 * 1000 kg * (115.83 m/s)^2
Force * 30 m = 1/2 * 1000 kg * 13418.96 m^2/s^2
Force * 30 = 6714480 N
Force ≈ 223816 N (rounded to the nearest whole number)
So, the magnitude of the average force acting on the car during its deceleration is approximately 223,816 Newtons.